limit

[canvas]

can you tell me guys if am I right?$$\lim_{x\to0}x\sin\left(\frac{\pi}{x}\right)=\lim_{x\to0}x\frac{\sin\left(\frac{\pi}{x}\right)}{\frac{\pi}{x}}\frac{x}{\pi}=\pi\lim_{x\to0}\frac{\sin\left(\frac{\pi}{x}\right)}{\frac{\pi}{x}}\lim_{x\to0}x^2=\pi\cdot1\cdot0=0$$

 

[canvas]

That is not correct!

Do you know:

$\displaystyle \lim_{x \to 0} \frac{sin(x)}{x} \ = ?$

$$\lim_{x\to0}\frac{\sin(x)}{x}=1$$
what's wrong sir?

 

[pka]

can you tell me guys if am I right?$$\lim_{x\to0}x\sin\left(\frac{\pi}{x}\right)=\lim_{x\to0}x\frac{\sin\left(\frac{\pi}{x}\right)}{\frac{\pi}{x}}\frac{x}{\pi}=\pi\lim_{x\to0}\frac{\sin\left(\frac{\pi}{x}\right)}{\frac{\pi}{x}}\lim_{x\to0}x^2=\pi\cdot1\cdot0=0$$

Study this LINK and tell us what you think.

 

[Deleted member 4993]

$$\lim_{x\to0}\frac{\sin(x)}{x}=1$$
what's wrong sir?

Your numerical result is CORRECT - I was mistaken.

 

[canvas]

ok, I think that this is the mistake I've made:

$$\lim_{x\to0}\frac{\sin(\square)}{\square}=1\,\,\Leftrightarrow\,\,\lim_{x\to0}\square=0$$
am I right?

 

[pka]

ok, I think that this is the mistake I've made:
$$\lim_{x\to0}\frac{\sin(\square)}{\square}=1\,\,\Leftrightarrow\,\,\lim_{x\to0}\square=0$$am I right?

Actually NO. You have missed that $\mathop {\lim }\limits_{x \to 0} \left( {\frac{\pi }{x}} \right)\ne 0$
$\left| {x\sin \left( {\dfrac{\pi }{x}} \right)} \right| = \left| x \right|\left| {\sin \left( {\dfrac{\pi }{x}} \right)} \right| \leqslant \left| x \right|$ Can you explain that?
What is the limit?

 

[Steven G]

ok, I think that this is the mistake I've made:

$$\lim_{x\to0}\frac{\sin(\square)}{\square}=1\,\,\Leftrightarrow\,\,\lim_{x\to0}\square=0$$
am I right?

I am assuming that you are saying that the same expressions go into the first two boxes. OK, that is fine.
In $\displaystyle \lim_{x\to0}\dfrac{sinx}{x}=1$, note that not only are the two x's exactly the same but they are both going to 0 as x goes to 0.
In $\displaystyle \lim_{x\to0}\dfrac{sin2x}{2x}=1$, note that not only are the two 2x's exactly the same but they are both going to 0 as x goes to 0.
In $\displaystyle \lim_{x\to5}\dfrac{sin(x-5)}{x-5}=1$, note that not only are the two (x-5)'s exactly the same but they are both going to 0 as x goes to 5
In $\displaystyle \lim_{x\to\infty}\dfrac{sin(1/x)}{1/x}=1$, note that not only are the two (1/x)'s exactly the same but they are both going to 0 as x goes to $\displaystyle \infty$.

 

[Dr.Peterson]

can you tell me guys if am I right?$$\lim_{x\to0}x\sin\left(\frac{\pi}{x}\right)=\lim_{x\to0}x\frac{\sin\left(\frac{\pi}{x}\right)}{\frac{\pi}{x}}\frac{x}{\pi}=\pi\lim_{x\to0}\frac{\sin\left(\frac{\pi}{x}\right)}{\frac{\pi}{x}}\lim_{x\to0}x^2=\pi\cdot1\cdot0=0$$

One error is in your algebra. It is not true that $\sin\left(\frac{\pi}{x}\right)=\frac{\sin\left(\frac{\pi}{x}\right)}{\frac{\pi}{x}}\frac{x}{\pi}$. Can you see the error?

Of course, the bigger error is what @pka pointed out, namely that $\lim_{x\to0}\frac{\sin\left(\frac{\pi}{x}\right)}{\frac{\pi}{x}}\ne1$ because $\lim_{x\to0}\frac{\pi}{x}\ne0$

 

[Steven G]

The reason why $\displaystyle \lim_{x\to 0}\dfrac{sinx}{x} = 1$ is because sinx and x are extremely close to one another when x is near 0. As you know, when you divide two numbers that are close to one another you get a result close to 1.
As mentioned in my earlier post, not only do the two arguments have to be the same for the limit to be 1 but also the arguments have to approach 0 as x goes to whatever it is going to.

 

[canvas]

so it should be: $$\lim_{x\to\triangle}\frac{\sin(\square)}{\square}=1\,\,\Leftrightarrow\,\,\lim_{x\to\triangle}\square=0$$ right?

 

[Steven G]

so it should be: $$\lim_{x\to\triangle}\frac{\sin(\square)}{\square}=1\,\,\Leftrightarrow\,\,\lim_{x\to\triangle}\square=0$$ right?

You seem to be understanding what I said. Good!